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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1901_2000.s1929_concatenation_of_array;

// #Easy #Array #2022_05_15_Time_1_ms_(92.52%)_Space_42.9_MB_(80.72%)

/**
 * 1929 - Concatenation of Array\.
 *
 * Easy
 *
 * Given an integer array `nums` of length `n`, you want to create an array `ans` of length `2n` where `ans[i] == nums[i]` and `ans[i + n] == nums[i]` for `0 <= i < n` ( **0-indexed** ).
 *
 * Specifically, `ans` is the **concatenation** of two `nums` arrays.
 *
 * Return _the array_ `ans`.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,2,1]
 *
 * **Output:** [1,2,1,1,2,1]
 *
 * **Explanation:** The array ans is formed as follows: 
 *
 * - ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]] 
 *
 * - ans = [1,2,1,1,2,1]
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,3,2,1]
 *
 * **Output:** [1,3,2,1,1,3,2,1]
 *
 * **Explanation:** The array ans is formed as follows: 
 *
 * - ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]] 
 *
 * - ans = [1,3,2,1,1,3,2,1]
 *
 * **Constraints:**
 *
 * *   `n == nums.length`
 * *   `1 <= n <= 1000`
 * *   `1 <= nums[i] <= 1000`
**/
public class Solution {
    public int[] getConcatenation(int[] nums) {
        int[] result = new int[nums.length * 2];
        System.arraycopy(nums, 0, result, 0, nums.length);
        int i = nums.length;
        for (int j = 0; i < result.length && j < nums.length; i++, j++) {
            result[i] = nums[j];
        }
        return result;
    }
}